When we transform $a\sin x+b\cos x=c$ into $a\sin x+b\cos x=R\sin(x+k)$, we calculate the $k$ angle by $k=\tan(b/a)$. By using calculator, we get a positive or negative degree value for $k$. I know that when we invert $\tan$, we get many solutions, due to the periodic nature of the tan function, which are given by $180n+m$ (where n is an integer) and $m$ is the value of $k$ that the calculator gave us. Then, we somehow determine which of the $180n+m$ we take for $k$ (based on the signs of $a$ and $b$). My question is why we do this? I assume that since we are talking about a function with period, the $R\sin(x+k)$, and the $k$ is a shift of the $\sin x$ function across the x line, we need to specify which $k$ of the $180n+m$ we take.
Can you elaborate please? Thanks
1 Answer
Let's take an arbitrary equation $$ A\sin(x) + B\cos(x) = R\sin(x + \phi) \quad (1) $$ We can write $\sin(x + \phi)$ as $$ \sin(x + \phi) = \sin(x)\cos(\phi) + \sin(\phi)\cos(x)\qquad (2) $$ Therefore, $R\sin(x + \phi) = R\sin(x)\cos(\phi) + R\sin(\phi)\cos(x)$. Let's equate equations (1) and (2). $$ A\sin(x) + B\cos(x) = R\sin(x)\cos(\phi) + R\sin(\phi)\cos(x) $$ So $A = R\cos(\phi)$ and $B = R\sin(\phi)$. Then $A^2 + B^2 = R^2(\cos^2(\phi) + \sin^2(\phi)) = R^2$; that is, $R = \sqrt{A^2+B^2}$. Then $$ \tan(\phi) = \frac{B\sin(\phi)}{A\cos(\phi)} =\frac{B}{A}\\ \phi = \tan^{-1}\Big(\frac{B}{A}\Big) $$ where if $A,B > 0$ we are in quadrant 1, $A>0$ and $B<0$ we are in quad 4, $A<0$ and $B>0$ we are in quad 2, and quad 3 for $A,B<0$.
We can then do this for $R\cos(x-\phi)$ as well where $$ R\cos(x-\phi) = R\cos(x)\cos(\phi) + R\sin(x)\sin(\phi) $$ Why dont you give it a try.
